Work out 35% of 260 [2]
- for a correct method, e.g. $10\% = 26$ and $5\% = 13$, or $0.35 \times 260$, or $3 \times 26 + 13$M1
- $91$A1
- Answer: 91
- Examiner tip: Build 35% from 10% (×3) plus 5% (half of 10%) and show each step.
(a) A coat normally costs £84. In a sale, the price is reduced by 15%. Work out the sale price of the coat. [2]
(b) Write 18 as a percentage of 75. [2]
- (a) for $84 \times 0.85$ or $84 - 84 \times 0.15$ (or $12.6$ seen)M1
- (a) £71.40A1
- (b) for $\frac{18}{75}$ or $18 \div 75 \times 100$M1
- (b) $24\%$A1
- Answer: (a) £71.40 (b) 24%
- Examiner tip: Money answers need two decimal places: £71.4 may lose the accuracy mark.
Hamid bought a house for £185 000. He sold the house for £212 750.
Work out his percentage profit. [3]
- for $212\,750 - 185\,000 = 27\,750$ or $\frac{212\,750}{185\,000} = 1.15$M1
- for $\frac{27\,750}{185\,000} \times 100$ or $(1.15 - 1) \times 100$M1
- $15\%$A1
- Answer: 15%
- Examiner tip: Percentage change is always divided by the ORIGINAL value (£185 000), not the new one.
The price of a bike is increased by 12%. The new price is £403.20
Work out the price of the bike before the increase. [3]
- for recognising $112\%$ represents £403.20 or multiplier $1.12$M1
- for $403.20 \div 1.12$ or $403.20 \div 112 \times 100$M1
- £360A1
- Answer: £360
- Examiner tip: Do not take 12% off £403.20 (giving £354.82) – the 12% was of the original price, so divide by 1.12.
Ravi needs to buy 6 tins of paint. He sees these offers.
| Shop A £12.40 per tin 20% off everything | Shop B £11.20 per tin 15% off when you buy 5 or more tins |
Which shop is cheaper for 6 tins, and by how much? You must show your working. [4]
- for a method to find the Shop A cost, e.g. $6 \times 12.40 \times 0.8$ (or $74.40$ and $14.88$ seen)M1
- for a method to find the Shop B cost, e.g. $6 \times 11.20 \times 0.85$ (or $67.20$ and $10.08$ seen)M1
- for £59.52 and £57.12A1
- Shop B is cheaper by £2.40 (ft their two totals if both M marks awarded)C1
- Answer: Shop B, by £2.40 (A £59.52, B £57.12)
- Examiner tip: A comparison question needs a clear statement at the end ('Shop B is cheaper by £2.40') supported by both totals.
The price of a phone is increased by 20%. The new price is then decreased by 20%.
Sam says, 'The phone is now back to its original price.'
Is Sam correct? You must show how you get your answer. [2]
- for $1.2 \times 0.8$ or a worked example with any original price, e.g. £100 → £120 → £96M1
- No, with correct supporting statement, e.g. multiplier is $0.96$ so the price is 4% lower than the originalC1
- Answer: No – overall multiplier 0.96, a 4% decrease
- Examiner tip: The second 20% is of a bigger amount, so the decrease is larger than the increase; 'No' alone scores 0.
In a sale, all prices are reduced by 30%. Jo also has a voucher that takes a further 10% off the sale price.
Jo pays £50.40 for a jacket.
(a) Work out the original price of the jacket. [3]
(b) Jo says, 'Altogether that is 40% off the original price.' Explain why Jo is wrong. [1]
- (a) for $0.7\times 0.9$ $(=0.63)$ oe (e.g. 63%), or $50.40 \div 0.9$ $(=56)$, or $50.40 \div 0.7$ $(=72)$M1
- (a) for a complete method, e.g. $50.40 \div (0.7 \times 0.9)$ or $50.40 \div 0.63$ or $56 \div 0.7$ or $72 \div 0.9$ or $50.40 \div 63 \times 100$M1
- (a) £80A1
- (b) e.g. the 10% is taken off the reduced (sale) price, so the overall multiplier is $0.7 \times 0.9 = 0.63$, which is a 37% reduction (not 40%); allow a correct counter-example, e.g. £100 → £70 → £63, or using (a): £80 − 40% = £48, not £50.40; allow a correct reason without numbers, e.g. '10% of the sale price is less than 10% of the original price'. Do not accept '37%' alone with no reason or workingC1
- Answer: (a) £80 (b) overall multiplier 0.63, i.e. 37% off
- Examiner tip: Reverse percentages: divide by the multiplier – never add 30% and 10% back on to £50.40.
(a) Write 0.07 as a percentage. [1]
(b) Write these numbers in order of size. Start with the smallest number.
$62\%\qquad 0.6\qquad \dfrac{13}{20}\qquad 0.615$
[2]
- (a) 7% (accept 7.0%)B1
- (b) for converting at least two of the values to a common form, e.g. 0.62 and 0.65, or 60% and 61.5%, or $\frac{13}{20} = \frac{65}{100}$M1
- (b) $0.6,\ 0.615,\ 62\%,\ \frac{13}{20}$ (accept in converted form, e.g. 0.6, 0.615, 0.62, 0.65; accept in reverse order only if clearly labelled 'largest first')A1
- Answer: (a) 7% (b) 0.6, 0.615, 62%, 13/20
- Examiner tip: 0.07 is 7 hundredths, so 7% – not 70% or 0.7%; and 0.615 is bigger than 0.6 but smaller than 0.62.
Priya puts £600 into a savings account.
The account pays 3% simple interest per year.
Work out the total amount of money in the account at the end of 4 years. [3]
- for finding 3% of 600, e.g. $600 \div 100 \times 3$ $(=18)$M1
- for multiplying their interest for one year by 4 $(=72)$, or $12\%$ of 600 seenM1
- £672A1
- Answer: £672
- Examiner tip: Simple interest is the same every year – work out one year's interest, multiply by the number of years, then add it on to the £600.
Ella took a French test and a Spanish test.
In French she scored 34 out of 40.
In Spanish she scored 52 out of 64.
In which test did Ella get the better result? You must show how you get your answer. [3]
- for $\frac{34}{40}\times 100$ oe, or 85% or 0.85 seenM1
- for $\frac{52}{64}\times 100$ oe, or 81.25% or 0.8125 seen (accept 81.3% or 81%)M1
- French, with 85% and 81.25% (or equivalent comparable values, e.g. 0.85 and 0.8125, or both scores scaled to the same total) – the decision must be supported by correct figuresC1
- Answer: French (85% compared with 81.25%)
- Examiner tip: Compare like with like – change both scores to percentages (or decimals) and then write a clear conclusion naming the test.
The bar chart shows the number of members of a running club at the start of four months in one year.
(a) Work out the percentage increase in the number of members from January to July. [3]
(b) Work out the percentage decrease in the number of members from July to October.
Give your answer correct to 1 decimal place. [2]
- (a) reads January = 160 and July = 220 from the chart (may be implied by 60 seen)B1
- (a) for $\frac{220-160}{160}\times 100$ oe, or $\frac{220}{160} = 1.375$ and $-1$ (or $-100\%$) ft their readingsM1
- (a) 37.5(%)A1
- (b) for $\frac{220-190}{220}\times 100$ oe, or $100 - \frac{190}{220}\times 100$ (ft their July reading); do not award for dividing by 190M1
- (b) 13.6(%) (from 13.636…)A1
- Answer: (a) 37.5% (b) 13.6%
- Examiner tip: Percentage change is always divided by the ORIGINAL value – for July to October that is 220, not 190.
In a sale, all prices are reduced by 25%.
The sale price of a lamp is £36.
Work out the price of the lamp before the sale. [3]
- for recognising £36 represents 75%, e.g. $36 \div 75$ $(=0.48)$ or $36 \div 3$ $(=12)$ for 25%M1
- for a complete method, e.g. $36 \div 75 \times 100$ or $12 \times 4$ or $36 \div 0.75$ or $36 + 12$M1
- £48A1
- Answer: £48
- Examiner tip: Don't add 25% of £36 back on (that gives £45) – £36 is 75% of the original price, so find 1% or 25% first.
The population of a town in 2025 was 238% of its population in 1975.
The population in 2025 was 59 500.
(a) Work out the population of the town in 1975. [2]
(b) Write down the percentage increase in the population from 1975 to 2025. [1]
- (a) for $59\,500 \div 2.38$ or $59\,500 \div 238 \times 100$ oeM1
- (a) 25 000A1
- (b) 138(%)B1
- Answer: (a) 25 000 (b) 138%
- Examiner tip: 238% of the old value means multiply by 2.38 – the increase is 238% − 100% = 138%, not 238%.
60% of the students at a school are girls. The rest are boys.
35% of the girls walk to school.
45% of the boys walk to school.
(a) Work out the percentage of all the students at the school who walk to school. [2]
156 students walk to school.
(b) Work out the total number of students at the school. [2]
- (a) for $0.6 \times 0.35$ $(=0.21)$ or $0.4 \times 0.45$ $(=0.18)$ oe; or uses a chosen total, e.g. out of 100 students: 60 girls → 21, 40 boys → 18M1
- (a) 39(%)A1
- (b) for $156 \div 39 \times 100$ or $156 \div 0.39$ oe, ft their (a), e.g. 39% → 156 so 1% → 4M1
- (b) 400A1
- Answer: (a) 39% (b) 400
- Examiner tip: You can't average 35% and 45% to get 40% – there are more girls than boys, so work out each group as a percentage of the whole school.
A shop increases the price of a television by $x\%$.
Later, the shop decreases the new price by $x\%$.
The final price of the television is 9% less than the original price.
Work out the value of $x$. [4]
- for a multiplier for each change, e.g. $\left(1+\frac{x}{100}\right)$ and $\left(1-\frac{x}{100}\right)$, or using a price $P$: $P\left(1+\frac{x}{100}\right)$ then $\times\left(1-\frac{x}{100}\right)$M1
- for forming a correct equation, e.g. $\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right) = 0.91$ oe (e.g. $(100+x)(100-x) = 9100$)M1
- for simplifying to $1 - \frac{x^2}{10\,000} = 0.91$ or $x^2 = 900$ oeM1
- 30 (ignore −30 seen with 30 if 30 chosen; a correct trial, e.g. $1.3 \times 0.7 = 0.91$, with no algebra scores full marks only if 30 is clearly stated as the answer)A1
- Answer: $x = 30$
- Examiner tip: The decrease is x% of the NEW (bigger) price, so the changes don't cancel – use multipliers and multiply them, never add them.
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